# Wilson loop as path integral of parallel transport action

I am trying to get that the path integral of the parallel transport action is the Wilson loop. Here is the setting:

Let $$w$$ be a complex vector dimension $$N$$, and $$A_{\mu}$$ a fixed Yang-Mills connection. We will work with $$G=SU(N)$$. Using the parallel transport equation and the constraint:

$$i\frac{dw}{d\tau} = \frac{dx^{\mu}}{d\tau} A_{\mu}w$$ $$w^{\dagger}w=1$$ We construct the following action with a Lagrange multiplier $$\lambda$$ enforcing the constraint above. Evidently the equation of motion of this action is the parallel transport equation. $$S_w = \int \left(i w^{\dagger}\frac{dw}{dt} + \lambda(w^{\dagger}w-1)+w^{\dagger}A(x(\tau))w \right)d\tau$$

This vector satisfies: $$[w_i,w_j^{\dagger}]=\delta_{ij}$$

Now let $$\tau \in \mathbf{S}$$ to allow for large gauge transformations. I finally arrive to the following path integral, where I had to insert the $$w_i$$ factors for it to not vanish:

$$Z_w[A]= \int e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty) \mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}$$

I am supposed to get that

$$Z_w[A] = tr \mathcal{P}e^{i\int A d\tau}$$

How do I compute that specific path integral?

• I think, that commutation equation, that you wrote, are in canonical quantization. But you are interested in path integral quantization of such system, right? Apr 6, 2020 at 16:22
• Yes. I was providing background for both quantizations, but mainly I'm interested in path integral. Apr 6, 2020 at 17:09
• Is it obvious, that path integral vanishes, if one didn't inserted $w_i$ factors? Apr 6, 2020 at 19:21

I will follow Monopoles and Wilson Lines, by David Tong, Kenny Wong.

Let's work in perturbation theory. Action: $$S_w = \int \left(i w^{\dagger}\frac{dw}{dt} + \alpha(w^{\dagger}w-\kappa)+w^{\dagger}A(x(\tau))w \right)d\tau$$ Propogator for free (non-interacting) part of action, in following we use $$\langle \dots \rangle$$ for averaging by free theory: $$\langle w_i^\dagger(\tau_1) w_j(\tau_2)\rangle = \theta(\tau_1-\tau_2) \delta_{ij}$$

1. Vacuum bubbles

As usual in perturbation theory, we can factorise vacuum diagrams. You will obtain up to notations:

All $$n ≥ 2$$ factors vanish because the product of the step functions vanishes everywhere except on a set of measure zero. So we have only one contribution ($$\theta(0)=1/2$$): $$\exp\left(i (N/2 - \kappa )\int dt\; \alpha(t)\right) = \exp\left(-i \kappa_{eff} \int dt\; \alpha(t)\right)$$

1. Path integral with insertions

$$Z_w[A]= \int\mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}\; e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty)$$

This integral correspond to following series of diagrams (times to vacuum bubbles factor):

This series correspondence to:

Including vacuum bubbles we left with:

$$Z_\omega[A] = W[A]\int D\alpha e^{-i\int dt \;\alpha(t) (\kappa_{eff}-1)}$$

If $$\kappa_{eff}= 1$$, we obtain:

$$Z_\omega[A] = W[A]$$

• Could you summarize why without insertion of the $w$ and $w^{\dagger}$ the path integral vanishes? Also, what would be the difference if there is no $\lambda$ apart from not conserving the norm of $w$? Apr 7, 2020 at 16:42
• Without insertion you will have only $\int D\alpha e^{-i\int dt \alpha(t)\kappa_{eff}}= \delta(\kappa_{eff})$. Apr 7, 2020 at 18:58
• What about without imposing $\alpha$? Apr 7, 2020 at 19:08
• I don't understand your question. $\alpha$ is present in action from initial step. Apr 7, 2020 at 20:26
• Yes, formally this integral is infinity, so you need be ore careful when deal with correlators. You need divide integral by average of identity operator. Infinity will be cancel. Apr 8, 2020 at 1:42